gaurav.s23 Ответов: 1

Как скачать файл с сервера в веб-методе ?


I do not know how to achieve to download file from server using web method. I tried something like this. 
I passed data and file name from angularJS to my web method, it creates excel file and upload in server. 
after completion of file creation and upload in server i returned url to another webform on success call to my JS function. 
From there I tried to open new web form where i am trying to implement code to download file. But how would my code will know which life to 
download since there could be multiple file presented. how so i pass fileName as well so that i could identified the file to download.

Please check my code and let me know what to change and what to do ? or anyother way to achieve this. 
Thank you.


Что я уже пробовал:

[WebMethod]
        public static string ExportExcel(List<Student> students, string fileName)
        {
          
                    ListtoDataTableConverter converter = new ListtoDataTableConverter();

                    DataTable dt = converter.ToDataTable(students);
                    DataSet ds = new DataSet();
                    ds.Tables.Add(dt);

                    string FilePath = fileLocation + fileName + DateTime.Now.ToString("yyyyMMddHHmmssfff") + ".xlsx"; ;

                    DataSetToExcel.ExportDataSet(FilePath, ds);

                    HttpContext.Current.Response.ContentType = "application/octet-stream";
                    HttpContext.Current.Response.AppendHeader("Content-Disposition", "filename=" + Path.GetFileName(FilePath));
                  
                    return JsonConvert.SerializeObject(new { Success = true, redirectUrl = VirtualPathUtility.ToAbsolute("~/DownloadExcelForm.aspx") });
       
        }
		
		

Javascript:

  $scope.SendFile = function (filteredItems) {
        var xlFile = $scope.xlFilename;
        var students = [];
        students = filteredItems ;
      
        $http({
            method: "POST",
            url: "ReportStudent.aspx/ExportExcel",
            data: { students: students, fileName: xlFile },
            contentType: "application/json",
            dataType: "json"
        }).then(function (response) {
            var responsedata = JSON.parse(response.data.d);
            window.location.href = responsedata.redirectUrl;
          
        });

    };
	
Webform Pageload:
 protected void Page_Load(object sender, EventArgs e)
        {
           
            string filepath = "";
            Response.ContentType = ContentType;
          //  Response.AppendHeader("Content-Disposition", "attachment; filename=" + Path.GetFileName(FilePath)); how to pass this file path ????
            Response.TransmitFile(FilePath);
            Response.Flush();
            System.Threading.Thread.Sleep(1000);
                System.Threading.Thread.Sleep(1000);
                if (File.Exists(filepath))
                {
                    File.Delete(filepath);
                }
                //response.writefile(filepath)
                Response.End();
            
        }

1 Ответов

Рейтинг:
1

Zahid Mughal

[WebMethod]
public byte[] DownloadFile(string FName)
{
    System.IO.FileStream fs1=null;
    fs1=System.IO.File.Open(FName,FileMode.Open,FileAccess.Read);
    byte[] b1=new byte[fs1.Length];
    fs1.Read(b1,0,(int)fs1.Length);
    fs1.Close();
    return b1; 
}


Используйте следующий справочный код, надеюсь, он будет полезен для вас.


[no name]

Я добился своего передаваемого имени файла в строке запроса.

[no name]

Спасибо, что пришли мне на помощь. Я бы тоже попробовал ссылку на код и дал вам знать.

[no name]

Добро пожаловать и, пожалуйста, не забудьте отметить мой хороший рейтинг. :)